ANSWERS TO CHAPTER 1 HOMEWORK PROBLEMS
1-1. The reaction for the formation of Al(OH)3(s) from alum is:
Al2(SO4)3 • 14 H2O → 2 Al(OH)3(s) + 3 SO42− + 6 H+ + 8 H2O
The reaction could be written in a few different ways, but the important relationship is that two
moles of Al(OH)3(s) are created per mole of alum added. The mass of alum required daily is:
( 60 mg/L ) ( 7500 m3 /d )(103 L/m3 )(1kg/106 mg ) = 450 kg alum/d
The molecular weight of alum is 594, and that of Al(OH)3(s) is 78, so if all the Al precipitates as
Al(OH)3(s), the amount of Al(OH)3(s) formed is:
⎛ mol Al(OH)3 ⎞⎛ mg Alum ⎞ ⎛ 1 mol Alum ⎞ ⎛ 78, 000 mg Al(OH)3 ⎞
⎜2 ⎟ ⎜ 60 ⎟⎜ ⎟⎜ ⎟
⎝ mol Alum ⎠ ⎝ L ⎠ ⎝ 594, 000 mg Alum ⎠ ⎝ mole Al(OH)3 ⎠
= 15.8 mg Al(OH)3(s)/L
The total concentration of solids in the water is the sum of 18 mg/L solids in the water initially
and the 15.8 mg Al(OH)3/L created by precipitation, or 33.8 mg/L total suspended solids.
The daily load of particulate solids can then be computed as
⎛ mg ⎞ ⎛ m3 ⎞ ⎛ 3 L ⎞ ⎛ kg ⎞ kg
⎜ 33.8 ⎟ ⎜ 7500 ⎟ ⎜ 10 3 ⎟ ⎜ 1 6 ⎟ = 253.5
⎝ L ⎠⎝ d ⎠⎝ m ⎠ ⎝ 10 mg ⎠ d
1
,1-2. (a) Assuming that advection is the only significant mechanism for transporting the
contaminant and that the contaminant is non-reactive in the system, a mass balance written using
the entire water system as the control volume (shown as the dotted line in the figure below) is as
follows:
dmX
= ( Q − QR ) cmuni + X − ( Q − QR ) cout
dt
where mX is the mass of contaminant in the water system in the plant and cmuni is the contaminant
concentration in the incoming municipal water.
Q – QR Q – QR
UPW
cmuni cout
QR, cR
dmX
and cmuni are both zero, so the equation simplifies to one that yields the value of cout
dt
directly:
0 = ( Q − QR ) 0 + X − ( Q − QR ) cout
X
cout =
Q − QR
(b) When the plant is operating with the minimum usage of municipal water, the concentration of
contaminant entering the UPW plant is cacc. Under those conditions, the system schematic is as
shown below.
2
, X
Q – QR Q, cin=cacc Q – QR
UPW
cmuni cout
QR, cR
A mass balance written around the point at which the recycle stream mixes with the municipal
water (circled in the schematic) is as follows:
dmX
= ( Q − QR ) cmuni + QR cR − Qcacc
dt
Again applying the steady-state assumption, noting that cR equals cout, and substituting the
expression from part a for cout, the mass balance can be re-written and an expression for QR can
be derived as follows:
0 = ( Q − QR ) 0 + QR cout − Qcacc
Qcacc Qcacc
QR = =
cout ⎛ X ⎞
⎜ ⎟
⎝ Q − QR ⎠
Algebraic manipulation yields the following explicit expressions for QR and R:
Qcacc
QR = Q
X + Qcacc
QR Qcacc
R≡ =
Q
X + Qcacc
(c) The given information indicates that Q = 2500 L/min, X = 300 mg/min, and cacc = 0.2 mg/L.
Substituting these values into the expression derived in part (b), we find:
R=
( 2500 L/min )( 0.2 mg/L ) = 0.625
300 mg/min + ( 2500 L/min )( 0.2 mg/L )
3
,We can determine QR and Qmuni as:
⎛ L ⎞ L
QR = RQ = ( 0.625) ⎜ 2500 ⎟ = 1562
⎝ min ⎠ min
L L L
Qmuni = Q − QR = 2500 − 1562 = 938
min min min
4
, 1-3. (a) The problem can be solved by writing (separate) mass balances on water and suspended
solids. The control volume is taken as the entire reactor, and it can reasonably be assumed that
the influent and effluent structures are designed so that negligible dispersion occurs across the
boundaries. Since the density of water is constant, a mass balance on water becomes a volume
balance. With the reactor acting at steady state, the rate of change of storage of both water and
solids is zero. Water is not reactive, and the problem statement indicates that the solids also do
not participate in any reactions, so neither mass balance includes a reaction term. The two mass
balances can therefore be written as follows:
For water:
⎡ Rate of change of ⎤ ⎡ Rate of ⎤ ⎡ Rate of ⎤
⎢storage of water ⎥ ⎡ Rate of ⎤ ⎢ ⎥ ⎢ ⎥
⎥ = ⎢ water input ⎥ − ⎢
water output water output
⎢ ⎥−⎢ ⎥
⎢in the reactor ⎥ ⎢ ⎥ ⎢ by advection in ⎥ ⎢ by advection in ⎥
⎢ ⎥ ⎢⎣by advection ⎥⎦ ⎢ ⎥ ⎢ ⎥
⎣ with time ⎦ ⎣ the outlet stream ⎦ ⎣ the waste stream ⎦
0 = Qin − Qout − Qw
For suspended solids:
⎡ Rate of change of ⎤ ⎡ Rate of output ⎤ ⎡ Rate of output ⎤
⎢storage of suspended ⎥ ⎡ Rate of input ⎤ ⎢ ⎥ ⎢ of suspended solids⎥
⎥ = ⎢of suspended solids⎥ − ⎢
of suspended solids
⎢ ⎥−⎢ ⎥
⎢solids in the reactor ⎥ ⎢ ⎥ ⎢ by advection in ⎥ ⎢ by advection in ⎥
⎢ ⎥ ⎢⎣ by advection ⎥⎦ ⎢ ⎥ ⎢ ⎥
⎣ with time ⎦ ⎣ the outlet stream ⎦ ⎣ the waste stream ⎦
0 = Qin cin − Qout cout − Qw c w
From the water balance, we find Qout = Qin − Qw , and substituting that into the mass balance
for suspended solids yields:
Q in c in = (Q in − Q w )c out + Q w c w
= Q in c out − Q w c out + Q w c w
= Q in c out + Q w (c w − c out )
Rearranging, we find:
Qw ( cw − cout ) = Qin cin − Qin cout = Qin ( cin − cout )
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1-1. The reaction for the formation of Al(OH)3(s) from alum is:
Al2(SO4)3 • 14 H2O → 2 Al(OH)3(s) + 3 SO42− + 6 H+ + 8 H2O
The reaction could be written in a few different ways, but the important relationship is that two
moles of Al(OH)3(s) are created per mole of alum added. The mass of alum required daily is:
( 60 mg/L ) ( 7500 m3 /d )(103 L/m3 )(1kg/106 mg ) = 450 kg alum/d
The molecular weight of alum is 594, and that of Al(OH)3(s) is 78, so if all the Al precipitates as
Al(OH)3(s), the amount of Al(OH)3(s) formed is:
⎛ mol Al(OH)3 ⎞⎛ mg Alum ⎞ ⎛ 1 mol Alum ⎞ ⎛ 78, 000 mg Al(OH)3 ⎞
⎜2 ⎟ ⎜ 60 ⎟⎜ ⎟⎜ ⎟
⎝ mol Alum ⎠ ⎝ L ⎠ ⎝ 594, 000 mg Alum ⎠ ⎝ mole Al(OH)3 ⎠
= 15.8 mg Al(OH)3(s)/L
The total concentration of solids in the water is the sum of 18 mg/L solids in the water initially
and the 15.8 mg Al(OH)3/L created by precipitation, or 33.8 mg/L total suspended solids.
The daily load of particulate solids can then be computed as
⎛ mg ⎞ ⎛ m3 ⎞ ⎛ 3 L ⎞ ⎛ kg ⎞ kg
⎜ 33.8 ⎟ ⎜ 7500 ⎟ ⎜ 10 3 ⎟ ⎜ 1 6 ⎟ = 253.5
⎝ L ⎠⎝ d ⎠⎝ m ⎠ ⎝ 10 mg ⎠ d
1
,1-2. (a) Assuming that advection is the only significant mechanism for transporting the
contaminant and that the contaminant is non-reactive in the system, a mass balance written using
the entire water system as the control volume (shown as the dotted line in the figure below) is as
follows:
dmX
= ( Q − QR ) cmuni + X − ( Q − QR ) cout
dt
where mX is the mass of contaminant in the water system in the plant and cmuni is the contaminant
concentration in the incoming municipal water.
Q – QR Q – QR
UPW
cmuni cout
QR, cR
dmX
and cmuni are both zero, so the equation simplifies to one that yields the value of cout
dt
directly:
0 = ( Q − QR ) 0 + X − ( Q − QR ) cout
X
cout =
Q − QR
(b) When the plant is operating with the minimum usage of municipal water, the concentration of
contaminant entering the UPW plant is cacc. Under those conditions, the system schematic is as
shown below.
2
, X
Q – QR Q, cin=cacc Q – QR
UPW
cmuni cout
QR, cR
A mass balance written around the point at which the recycle stream mixes with the municipal
water (circled in the schematic) is as follows:
dmX
= ( Q − QR ) cmuni + QR cR − Qcacc
dt
Again applying the steady-state assumption, noting that cR equals cout, and substituting the
expression from part a for cout, the mass balance can be re-written and an expression for QR can
be derived as follows:
0 = ( Q − QR ) 0 + QR cout − Qcacc
Qcacc Qcacc
QR = =
cout ⎛ X ⎞
⎜ ⎟
⎝ Q − QR ⎠
Algebraic manipulation yields the following explicit expressions for QR and R:
Qcacc
QR = Q
X + Qcacc
QR Qcacc
R≡ =
Q
X + Qcacc
(c) The given information indicates that Q = 2500 L/min, X = 300 mg/min, and cacc = 0.2 mg/L.
Substituting these values into the expression derived in part (b), we find:
R=
( 2500 L/min )( 0.2 mg/L ) = 0.625
300 mg/min + ( 2500 L/min )( 0.2 mg/L )
3
,We can determine QR and Qmuni as:
⎛ L ⎞ L
QR = RQ = ( 0.625) ⎜ 2500 ⎟ = 1562
⎝ min ⎠ min
L L L
Qmuni = Q − QR = 2500 − 1562 = 938
min min min
4
, 1-3. (a) The problem can be solved by writing (separate) mass balances on water and suspended
solids. The control volume is taken as the entire reactor, and it can reasonably be assumed that
the influent and effluent structures are designed so that negligible dispersion occurs across the
boundaries. Since the density of water is constant, a mass balance on water becomes a volume
balance. With the reactor acting at steady state, the rate of change of storage of both water and
solids is zero. Water is not reactive, and the problem statement indicates that the solids also do
not participate in any reactions, so neither mass balance includes a reaction term. The two mass
balances can therefore be written as follows:
For water:
⎡ Rate of change of ⎤ ⎡ Rate of ⎤ ⎡ Rate of ⎤
⎢storage of water ⎥ ⎡ Rate of ⎤ ⎢ ⎥ ⎢ ⎥
⎥ = ⎢ water input ⎥ − ⎢
water output water output
⎢ ⎥−⎢ ⎥
⎢in the reactor ⎥ ⎢ ⎥ ⎢ by advection in ⎥ ⎢ by advection in ⎥
⎢ ⎥ ⎢⎣by advection ⎥⎦ ⎢ ⎥ ⎢ ⎥
⎣ with time ⎦ ⎣ the outlet stream ⎦ ⎣ the waste stream ⎦
0 = Qin − Qout − Qw
For suspended solids:
⎡ Rate of change of ⎤ ⎡ Rate of output ⎤ ⎡ Rate of output ⎤
⎢storage of suspended ⎥ ⎡ Rate of input ⎤ ⎢ ⎥ ⎢ of suspended solids⎥
⎥ = ⎢of suspended solids⎥ − ⎢
of suspended solids
⎢ ⎥−⎢ ⎥
⎢solids in the reactor ⎥ ⎢ ⎥ ⎢ by advection in ⎥ ⎢ by advection in ⎥
⎢ ⎥ ⎢⎣ by advection ⎥⎦ ⎢ ⎥ ⎢ ⎥
⎣ with time ⎦ ⎣ the outlet stream ⎦ ⎣ the waste stream ⎦
0 = Qin cin − Qout cout − Qw c w
From the water balance, we find Qout = Qin − Qw , and substituting that into the mass balance
for suspended solids yields:
Q in c in = (Q in − Q w )c out + Q w c w
= Q in c out − Q w c out + Q w c w
= Q in c out + Q w (c w − c out )
Rearranging, we find:
Qw ( cw − cout ) = Qin cin − Qin cout = Qin ( cin − cout )
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